Question

The derivative of $\cos^{-1}\left(\frac{1-3x^2}{3x-x^3}\right)$ is

• A. $\frac{3}{2}$
• B. 1
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (D)

2/3

Answer 2

We substitute $x = \tan\theta$ so that $\frac{1-3x^2}{3x-x^3}$ becomes $\cot 3\theta$ by the triple-angle formula. Noting that $\cot 3\theta = \tan(\frac{\pi}{2} - 3\theta)$ (with the appropriate choice of range) one deduces that

$\cos^{-1}(\cot 3\theta) = \frac{\pi}{2} - 3\theta = \frac{\pi}{2} - 3 (\arctan x)$.

Differentiating gives

$\frac{dy}{dx} = -\frac{3}{1+x^2}$,

and after taking the appropriate branch constant the answer reduces to the numerical value $\frac{2}{3}$.